Let U={1,2,3,4,5,6,7,8,9},A = {2,4,6,8} and B={2,3,5,7} . Verify that :
(i) `(AuuB)^(c)=A^(c)capB^(c)`
(ii) `(AcapB)^(c)=A^(c)uuB^(c)`.

Answer

To verify the given set identities, we will follow a systematic approach to find the required unions and intersections, as well as their complements.

### Given:
- Universal Set \( U = \{1, 2, 3, 4, 5, 6, 7, 8, 9\} \)
- Set \( A = \{2, 4, 6, 8\} \)
- Set \( B = \{2, 3, 5, 7\} \)

### (i) Verify that \( (A \cup B)^c = A^c \cap B^c \)

**Step 1: Find \( A \cup B \)**
To find \( A \cup B \), we take all distinct elements from both sets:
- From \( A \): 2, 4, 6, 8
- From \( B \): 2, 3, 5, 7

Combining these, we have:
\[ A \cup B = \{2, 3, 4, 5, 6, 7, 8\} \]

**Step 2: Find \( (A \cup B)^c \)**
The complement of \( A \cup B \) in the universal set \( U \):
\[ (A \cup B)^c = U - (A \cup B) = \{1, 9\} \]

**Step 3: Find \( A^c \)**
The complement of \( A \):
\[ A^c = U - A = \{1, 3, 5, 7, 9\} \]

**Step 4: Find \( B^c \)**
The complement of \( B \):
\[ B^c = U - B = \{1, 4, 6, 8, 9\} \]

**Step 5: Find \( A^c \cap B^c \)**
Now, we find the intersection of \( A^c \) and \( B^c \):
- \( A^c = \{1, 3, 5, 7, 9\} \)
- \( B^c = \{1, 4, 6, 8, 9\} \)

The common elements are:
\[ A^c \cap B^c = \{1, 9\} \]

**Step 6: Compare \( (A \cup B)^c \) and \( A^c \cap B^c \)**
We have:
- \( (A \cup B)^c = \{1, 9\} \)
- \( A^c \cap B^c = \{1, 9\} \)

Thus, \( (A \cup B)^c = A^c \cap B^c \) is verified.

### (ii) Verify that \( (A \cap B)^c = A^c \cup B^c \)

**Step 1: Find \( A \cap B \)**
To find \( A \cap B \), we take the common elements from both sets:
- Common element: 2

Thus,
\[ A \cap B = \{2\} \]

**Step 2: Find \( (A \cap B)^c \)**
The complement of \( A \cap B \):
\[ (A \cap B)^c = U - (A \cap B) = \{1, 3, 4, 5, 6, 7, 8, 9\} \]

**Step 3: Find \( A^c \cup B^c \)**
Now we find the union of \( A^c \) and \( B^c \):
- \( A^c = \{1, 3, 5, 7, 9\} \)
- \( B^c = \{1, 4, 6, 8, 9\} \)

Combining these, we have:
\[ A^c \cup B^c = \{1, 3, 4, 5, 6, 7, 8, 9\} \]

**Step 4: Compare \( (A \cap B)^c \) and \( A^c \cup B^c \)**
We have:
- \( (A \cap B)^c = \{1, 3, 4, 5, 6, 7, 8, 9\} \)
- \( A^c \cup B^c = \{1, 3, 4, 5, 6, 7, 8, 9\} \)

Thus, \( (A \cap B)^c = A^c \cup B^c \) is verified.

### Conclusion:
Both identities are verified:
1. \( (A \cup B)^c = A^c \cap B^c \)
2. \( (A \cap B)^c = A^c \cup B^c \)